Calculate the pH of 0.0048 M Ba(OH)2
Use this premium chemistry calculator to find hydroxide concentration, pOH, and pH for barium hydroxide and similar strong bases. The default example is set to 0.0048 M Ba(OH)2, which dissociates to release two hydroxide ions per formula unit.
Strong Base pH Calculator
Expert Guide: How to Calculate the pH of 0.0048 M Ba(OH)2
To calculate the pH of 0.0048 M Ba(OH)2, you need to recognize that barium hydroxide is a strong base and, in typical general chemistry problems, it is treated as fully dissociated in water. That matters because the pH is not found directly from the molarity of Ba(OH)2. Instead, you first determine the hydroxide ion concentration, written as [OH–], and then use the logarithmic relationship for pOH. Finally, you convert pOH to pH.
The key idea is stoichiometry. Each formula unit of Ba(OH)2 produces one Ba2+ ion and two OH– ions when dissolved:
Because of that 2 in front of OH–, a 0.0048 M solution of barium hydroxide produces twice that concentration of hydroxide ions. So the hydroxide concentration becomes:
Now calculate pOH using the base-10 logarithm:
At 25 C, use the common relationship:
So, the pH of 0.0048 M Ba(OH)2 is approximately 11.98. In most class settings, reporting the answer as pH = 11.98 or pH ≈ 12.0 is acceptable depending on the instructor’s rounding rules.
Step-by-step breakdown
- Identify Ba(OH)2 as a strong base.
- Write the dissociation equation: Ba(OH)2 → Ba2+ + 2OH–.
- Multiply the base molarity by 2 to find [OH–].
- Use pOH = -log[OH–].
- Use pH = 14 – pOH at 25 C.
Why the factor of 2 is essential
Students often miss the most important stoichiometric detail in this problem: Ba(OH)2 contains two hydroxide groups. If you accidentally use 0.0048 M as the hydroxide concentration, you will compute a pOH that is too large and a pH that is too small. That is the single most common mistake on this type of question.
Compare the incorrect and correct setup:
- Incorrect: [OH–] = 0.0048 M
- Correct: [OH–] = 2 × 0.0048 = 0.0096 M
The numerical difference looks small at first, but because pH and pOH use logarithms, even modest concentration changes produce meaningful shifts in the reported pH. In chemistry, stoichiometry always comes before logarithms.
Worked example in full detail
Let us solve the exact problem from start to finish in standard textbook form.
- Given: 0.0048 M Ba(OH)2
- Dissociation: Ba(OH)2 → Ba2+ + 2OH–
- Hydroxide concentration: [OH–] = 2(0.0048) = 0.0096 M
- pOH: pOH = -log(0.0096) = 2.0177
- pH: pH = 14.0000 – 2.0177 = 11.9823
- Rounded: pH ≈ 11.98
What chemistry assumptions are being used?
When teachers assign this problem, they usually expect three standard assumptions:
- Ba(OH)2 behaves as a strong base and dissociates completely.
- The solution is dilute enough that concentration can stand in for activity.
- The temperature is 25 C, so pH + pOH = 14.00.
These are exactly the assumptions used in introductory chemistry. In more advanced physical chemistry, you might account for activity coefficients, ionic strength, and temperature dependence of the ionic product of water. However, for school, college intro chemistry, and most online homework systems, the simple strong-base method is correct.
Common mistakes to avoid
- Forgetting the 2 in Ba(OH)2: This is the biggest error.
- Taking the negative log of the base molarity instead of [OH–]: Always convert to hydroxide concentration first.
- Mixing up pOH and pH: Strong bases are easiest to solve through pOH first.
- Using natural log instead of base-10 log: pH and pOH always use log base 10.
- Rounding too early: Keep several digits until the final step.
Comparison table: pH of selected Ba(OH)2 concentrations
The table below shows how the pH changes as the molarity of Ba(OH)2 changes. These values are calculated with the same approach used for 0.0048 M. Notice how doubling or halving concentration does not change pH linearly because pH is logarithmic.
| Ba(OH)2 concentration (M) | [OH-] produced (M) | pOH | pH at 25 C |
|---|---|---|---|
| 0.0010 | 0.0020 | 2.6990 | 11.3010 |
| 0.0025 | 0.0050 | 2.3010 | 11.6990 |
| 0.0048 | 0.0096 | 2.0177 | 11.9823 |
| 0.0100 | 0.0200 | 1.6990 | 12.3010 |
| 0.0500 | 0.1000 | 1.0000 | 13.0000 |
Comparison table: strong bases and hydroxide yield
This second table helps explain why different bases produce different hydroxide concentrations even at the same formal molarity. The chemistry depends on how many OH– ions each formula unit contributes upon dissociation.
| Base | Idealized dissociation | OH- ions per formula unit | [OH-] from a 0.0048 M solution |
|---|---|---|---|
| NaOH | NaOH → Na+ + OH- | 1 | 0.0048 M |
| KOH | KOH → K+ + OH- | 1 | 0.0048 M |
| Ca(OH)2 | Ca(OH)2 → Ca2+ + 2OH- | 2 | 0.0096 M |
| Ba(OH)2 | Ba(OH)2 → Ba2+ + 2OH- | 2 | 0.0096 M |
Why pH changes logarithmically
pH and pOH are logarithmic scales. That means each whole-number change represents a tenfold change in hydrogen ion concentration or hydroxide ion concentration. This is why chemistry students must be comfortable with logs when solving acid-base problems. For example, a solution with pOH 2 has ten times the hydroxide concentration of a solution with pOH 3.
For your specific problem, [OH–] = 0.0096 M is just a bit below 0.01 M. Since the negative log of 0.01 is 2, the pOH should be just a bit above 2. That quick estimate helps verify that the exact pOH of 2.0177 is reasonable. Then a pH just under 12 also makes sense. Developing this kind of number sense is a powerful chemistry skill.
How this problem appears in classes and exams
Problems like “calculate the pH of 0.0048 M Ba(OH)2” show up in high school chemistry, AP Chemistry review, first-year college chemistry, nursing prerequisites, and engineering fundamentals. Instructors use this question to test several skills at once:
- Recognizing strong electrolytes
- Writing dissociation equations correctly
- Using stoichiometric coefficients in solution chemistry
- Applying logarithms to concentration data
- Rounding to a sensible number of significant figures
If you can solve this problem accurately, you are demonstrating more than memorization. You are connecting formula interpretation, equilibrium conventions, and mathematical analysis.
Advanced note: what about temperature and real solutions?
In rigorous chemistry, the statement pH + pOH = 14 is strictly tied to the ionic product of water at 25 C. At other temperatures, the sum changes slightly because water autoionizes differently. Likewise, real solutions can deviate from ideality, especially as ionic strength rises. In most educational settings, though, the concentration here is low enough and the problem style simple enough that the standard 25 C assumption is expected and accepted.
If you are studying beyond introductory chemistry, you may also encounter the idea that not every hydroxide-containing compound behaves as an ideal strong base under all conditions. But for Ba(OH)2 in standard classroom stoichiometric pH problems, complete dissociation is the right model.
Quick mental check for the final answer
Here is a fast way to sanity-check the answer without doing the entire calculation from scratch:
- 0.0048 M Ba(OH)2 gives about 0.01 M OH– because of the factor of 2.
- A 0.01 M OH– solution has pOH about 2.
- If pOH is about 2, then pH is about 12.
So any answer near 12 is plausible. An answer near 10 or near 14 should immediately look suspicious. This type of estimate is useful on tests because it helps you catch calculator slips.
Authoritative references for pH concepts
For additional reading on pH, water chemistry, and acid-base fundamentals, review these trusted public resources:
Bottom line
To calculate the pH of 0.0048 M Ba(OH)2, multiply the base molarity by 2 to get [OH–] = 0.0096 M, find pOH = 2.0177, and subtract from 14. The final result is pH = 11.98 at 25 C. If you remember that Ba(OH)2 releases two hydroxide ions, the rest of the solution becomes straightforward.